`dy/dx = y + 3` Solve the differential equation

Saturday, October 24, 2009

$\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}={y}+{3}$ Solve the differential equation

Recall that in solving simple first order "ordinary differential equation" (ODE), we may apply variable separable differential equation wherein:

$\displaystyle{N}{\left({y}\right)}{y}'={M}{\left({x}\right)}$

$\displaystyle{N}{\left({y}\right)}\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}={M}{\left({x}\right)}$

$\displaystyle{N}{\left({y}\right)}{\left.{d}{y}\right.}={M}{\left({x}\right)}{\left.{d}{x}\right.}$

Before we can work on the direct integration: int N(y) dy= int M(x) dx to solve for the general solution of a differential equation.

For the given first order ODE: $\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}={y}+{3}$ can be rearrange by cross-multiplication into:

$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{y}+{3}}}={\left.{d}{x}\right.}$

Apply direct integration on both sides: $\displaystyle\int\frac{{{\left.{d}{y}\right.}}}{{{y}+{3}}}=\int$ dx

For the left side, we consider u-substitution by letting:

$\displaystyle{u}={y}+{3}$ then $\displaystyle{d}{u}={\left.{d}{y}\right.}$

The integral becomes:

$\displaystyle\int\frac{{{\left.{d}{y}\right.}}}{{{y}+{3}}}=\int\frac{{{d}{u}}}{{{u}}}$

Applying basic integration formula for logarithm:

$\displaystyle\int\frac{{{d}{u}}}{{{u}}}={\ln}{\left|{u}\right|}$

Plug-in $\displaystyle{u}={y}+{3}$ on $\displaystyle{\ln}{\mid}{u}$ $\displaystyle{\mid}$ , we get:

$\displaystyle\int\frac{{{\left.{d}{y}\right.}}}{{{y}+{3}}}={\ln}{\left|{y}+{3}\right|}$

For the right side, we apply the basic integration: $\displaystyle\int{\left.{d}{x}\right.}={x}+{C}$

Combing the results from both sides, we get the general solution of the differential equation as:

$\displaystyle{\ln}{\left|{y}+{3}\right|}={x}+{C}$

$\displaystyle{y}={{e}}^{{{x}+{C}}}-{3}$

$\displaystyle{y}={C}{{e}}^{{x}}-{3}$

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Saturday, October 24, 2009

$\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}={y}+{3}$ Solve the differential equation

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Thomas Jefferson's election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?

Saturday, October 24, 2009

Solve the differential equation

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Thomas Jefferson&#39;s election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?

$\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}={y}+{3}$ Solve the differential equation

Thomas Jefferson's election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?